A horizontal line on a distance-time graph means the object is
2 answers:
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Answer:
F = 2 × 10⁻³ N
Explanation:
Given:
frequency, f = 2400 MHz
Height, h = 25cm = 0.25 m
Area of the base, A = 30 cm x 30 cm = 900 cm² = 0.09 m²
Energy of the microwave, E = 0.50 mJ = 0.5 x 10⁻³ J
Now, the time taken by the wave from top to the base, t = h/c
here, c is the speed of the light
thus,
t = 0.25/(3 x 10⁸) = 8.33 x 10⁻¹⁰ s
The radiation pressure = Intensity/c
now, the intensity is given as:
I = Power/ area
also,
Power = Energy/ time = 0.5 x 10⁻³ J/8.33 x 10⁻¹⁰ s = 600000 W
thus,
I = 600000 W/ 0.09 m² = 6666666.6 W/m²
substituting the value in the formula for pressure due to radiation, we have
= 6666666.6 W/m²/(3 x 10⁸)
also
pressure = Force/ area
thus,
Force/ area = 6666666.6 W/m²/(3 x 10⁸)
or
Force (F) = (6666666.6 W/m² × 0.09 m²)/(3 x 10⁸)
or
F = 2 × 10⁻³ N
Answer:
Explanation:
Answer:
0.632 m
Explanation:
let the equilibrium separation between the charges is d and the angle made by string with the vertical is θ.
According to the diagram,
d = L Sinθ + L Sinθ = 2 L Sinθ .....(1)
Let T be the tension in the string.
Resolve the components of T.
T Sinθ = k q1 q2 / d^2
T Sinθ = k q1 q2 / (2LSinθ)² .....(2)
T Cosθ = mg .....(3)
Dividing equation (2) by equation (3), we get
tanθ = k q1 q2 / (4 L² Sin²θ x mg)
tan θ Sin²θ = k q1 q2 / (4 L² m g)
For small value of θ, tan θ = Sin θ
So,
Sin³θ = k q1 q2 / (4 L² m g)
Sin³θ = (9 x 10^9 x 0.785 x 10^-6 x 1.47 x 10^-6) / (4 x 0.5 x 0.5 x 4.20 x 10^-3 x 9.8)
Sin³θ = 0.2523
Sinθ = 0.632
θ = 39.2 degree
So, the separation between the two charges, d = 2 x L x Sin θ
d = 2 x 0.5 x 0.632 = 0.632 m
